Consider the following method. public double calculate (double x) { return x + 1.5; } The following code segment calls the method calculate in the same class. Double d1 = new Double (7.5); System. out. println (calculate (d1)); What, if anything, is printed when the code segment is executed? A. 8.0

B. 8.5

C. 9

D. 9.0

E. Nothing is printed because the code does not compile. The actual parameter d1 passed to calculate is a Double, but the formal parameter x is a double.

Question

Computers and Technology

Maximillian Santos

9 October, 11:45

Consider the following method. public double calculate (double x) { return x + 1.5; } The following code segment calls the method calculate in the same class. Double d1 = new Double (7.5); System. out. println (calculate (d1)); What, if anything, is printed when the code segment is executed? A. 8.0

B. 8.5

C. 9

D. 9.0

E. Nothing is printed because the code does not compile. The actual parameter d1 passed to calculate is a Double, but the formal parameter x is a double.

+1

Answers (1)

Know the Answer?

Regan Chen · Answer 1

E (Nothing is printed because the code will not compile)

Explanation:

There is a difference between Double which is a wrapper class and double which is data type. Changing the line Double d1 = new Double (7.5) to Double d1 = new (7.5) or simple declaring double d1 = 7.5 would have giving a printout of 9.0.